Prove: If H is a finite group of even cardinality, the H contains an element h of order 2.
Problem: Suppose $\displaystyle \left|H\right|$ is even. Prove that at least one element of $\displaystyle H$ is of order two.
Proof: Suppose that $\displaystyle H$ contained no elements of order two. We can see then that $\displaystyle a\ne a^{-1}$ for all nontrivial elements of $\displaystyle H$. Therefore elements of this nature come in distinct pairs ($\displaystyle a,a^{-1}$) adding up all these elements will then give an even number. Then noting that since no nontrivial element of $\displaystyle H$ had order two we can see that the total number of elements in $\displaystyle H$ is the number of all elements whose order isn't two and the identity element. But this is an even number plus one, thus odd which is of course a contradiction.